Question: Solve for $x$ and $y$ using substitution. ${x-y = 3}$ ${x = 4y-3}$
Since $x$ has already been solved for, substitute $4y-3$ for $x$ in the first equation. ${(4y-3)}{- y = 3}$ Simplify and solve for $y$ $4y-3 - y = 3$ $3y-3 = 3$ $3y-3{+3} = 3{+3}$ $3y = 6$ $\dfrac{3y}{{3}} = \dfrac{6}{{3}}$ ${y = 2}$ Now that you know ${y = 2}$ , plug it back into $\thinspace {x = 4y-3}\thinspace$ to find $x$ ${x = 4}{(2)}{ - 3}$ $x = 8 - 3$ ${x = 5}$ You can also plug ${y = 2}$ into $\thinspace {x-y = 3}\thinspace$ and get the same answer for $x$ : ${x - }{(2)}{= 3}$ ${x = 5}$